AMC12 2007 A

AMC12 2007 A · Q22

AMC12 2007 A · Q22. It mainly tests Remainders & modular arithmetic, Digit properties (sum of digits, divisibility tests).

For each positive integer $n$, let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$

对于每个正整数 $n$，令 $S(n)$ 表示 $n$ 的各位数字之和。有多少个 $n$ 满足 $n + S(n) + S(S(n)) = 2007?$

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (D)

正确答案：（D）

Solution

For the sake of notation, let $T(n) = n + S(n) + S(S(n))$. Obviously $n<2007$. Then the maximum value of $S(n) + S(S(n))$ is when $n = 1999$, and the sum becomes $28 + 10 = 38$. So the minimum bound is $2007-38=1969$. We do casework upon the tens digit: Case 1: $196u \Longrightarrow u = 9$. Easy to directly disprove. Case 2: $197u$. $S(n) = 1 + 9 + 7 + u = 17 + u$, and $S(S(n)) = 8+u$ if $u \le 2$ and $S(S(n)) = 2 + (u-3) = u-1$ otherwise. Case 3: $198u$. $S(n) = 18 + u$, and $S(S(n)) = 9 + u$ if $u \le 1$ and $2 + (u-2) = u$ otherwise. Case 4: $199u$. But $S(n) > 19$, and $n + S(n)$ clearly sum to $> 2007$. Case 5: $200u$. So $S(n) = 2 + u$ and $S(S(n)) = 2 + u$ (recall that $n < 2007$), and $2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1$. Fourth solution. In total we have $4 \mathrm{(D)}$ solutions, which are $1977, 1980, 1983,$ and $2001$.

为方便记号，令 $T(n) = n + S(n) + S(S(n))$。显然 $n<2007$。此时 $S(n) + S(S(n))$ 的最大值在 $n = 1999$ 时取得，为 $28 + 10 = 38$。因此 $n$ 的下界为 $2007-38=1969$。对个位数字分类讨论： Case 1: $196u \Longrightarrow u = 9$。可直接验证不成立。 Case 2: $197u$。$S(n) = 1 + 9 + 7 + u = 17 + u$，且当 $u \le 2$ 时 $S(S(n)) = 8+u$，否则 $S(S(n)) = 2 + (u-3) = u-1$。 Case 3: $198u$。$S(n) = 18 + u$，且当 $u \le 1$ 时 $S(S(n)) = 9 + u$，否则 $S(S(n)) = 2 + (u-2) = u$。 Case 4: $199u$。但此时 $S(n) > 19$，且 $n + S(n)$ 显然 $> 2007$。 Case 5: $200u$。则 $S(n) = 2 + u$ 且 $S(S(n)) = 2 + u$（注意 $n < 2007$），于是 $2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1$。这是第四个解。总共有 $4 \mathrm{(D)}$ 个解，分别为 $1977, 1980, 1983,$ 和 $2001$。

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